STOKES PROBLEM WITH SEVERAL TYPES OF BOUNDARY CONDITIONS IN AN EXTERIOR DOMAIN

Electronic Journal of Differential Equations, Vol. 2013 2013, No. 196, pp. 1 28. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu STOKES PROBLEM WITH SEVERAL TYPES OF BOUNDARY CONDITIONS IN AN EXTERIOR DOMAIN CHÉRIF AMROUCHE, MOHAMED MESLAMENI Abstract. In this article, we solve the Stoes problem in an exterior domain of R 3, with non-standard boundary conditions. Our approach uses weighted Sobolev spaces to prove the existence, uniqueness of wea and strong solutions. This wor is based on the vector potentials studied in [7] for exterior domains, and in [1] for bounded domains. This problem is well nown in the classical Sobolev spaces W m,2 when is bounded; see [3, 4]. 1. Introduction and functional setting Let denotes a bounded open in R 3 of class C 1,1, simply-connected and with a connected boundary =, representing an obstacle and is its complement; i.e. = R 3 \. Then a unit exterior normal vector to the boundary can be defined almost everywhere on ; it is denoted by n. The purpose of this paper is to solve the Stoes equation in, with two types of non standard boundary conditions on : u + π = f and div u = χ in, 1.1 u n = g and curl u n = h n on, and u + π = f and div u = χ in, π = π 0, u n = g n on and u n dσ = 0. 1.2 Since this problem is posed in an exterior domain, our approach is to use weighted Sobolev spaces. Let us begin by introducing these spaces. A point in will be denoted by x = x 1, x 2, x 3 and its distance to the origin by r = x = x 2 1 + x 2 2 + x 2 3 1/2. We will use the weights ρ = ρr = 1 + r 2 1/2. For all m in N and all in Z, we define the weighted space W m,2 = {u D : λ N 3 : 0 λ m, ρr m+ λ D λ u L 2 }, 2000 Mathematics Subject Classification. 35J25, 35J50, 76M30. Key words and phrases. Stoes equations; exterior domain; weighted Sobolev spaces; vector potentials; inf-sup conditions. c 2013 Texas State University - San Marcos. Submitted July 28, 2013. Published September 3, 2013. 1

2 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 which is a Hilbert space with the norm m u W m,2 = 1/2, ρ m+ λ D λ u 2 L 2 λ =0 where L 2 denotes the standard norm of L 2. We shall sometimes use the seminorm u W m,2 = 1/2. ρ D λ u 2 L 2 λ =m In addition, it is established by Hanouzet in [11], for domains with a Lipschitzcontinuous boundary, that D is dense in W m,2 m,2. We set W as the m,2 adherence of D for the norm W m,2. Then, the dual space of W, denoting by W m,2, is a space of distributions. Furthermore, as in bounded domain, we have for m = 1 or m = 2, W = {v W, v = 0 on }, W 0 = {v W 0, v = v n = 0 on }, where v n is the normal derivative of v. As a consequence of Hardy s inequality, the following Poincaré inequality holds: for m = 0 or m = 1 and for all in Z there exists a constant C such that m,2 v W, v W m,2 C v W m,2 ; 1.3 m,2 i.e., the seminorm W m,2 is a norm on W equivalent to the norm W m,2. In the sequel, we shall use the following properties. For all integers m and in Z, we have n Z with n m 2, P n W m,2, 1.4 where P n denotes the space of all polynomials of three variables of degree at most n, with the convention that the space is reduced to zero when n is negative. Thus the difference m is an important parameter of the space W m,2. We denote by Pn the subspace of all harmonic polynomials of P n. Using the derivation in the distribution sense, we can define the operators curl and div on L 2. Indeed, let, denote the duality pairing between D and its dual space D. For any function v = v 1, v 2, v 3 L 2, we have for any ϕ = ϕ 1, ϕ 2, ϕ 3 D, curl v, ϕ = v curl ϕ dx = and for any ϕ D, div v, ϕ = v 1 ϕ 3 ϕ 2 + v 2 ϕ 1 ϕ 3 + v 3 ϕ 2 ϕ 1 dx, x 2 x 3 x 3 x 1 x 1 x 2 ϕ ϕ ϕ v grad ϕ dx = v 1 + v 2 + v 3 dx. x 1 x 2 x 3 We note that the vector-valued Laplace operator of a vector field v = v 1, v 2, v 3 is equivalently defined by v = graddiv v curl curl v. 1.5

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 3 This leads to the following definitions: Definition 1.1. For all integers Z, we define the space with the norm H 2 curl, = {v W ; curl v W +1 }, v H 2 curl, = Also we define the space with the norm Finally, we set with the norm X 2 = v 2 W + curl v 2 W +1 1/2. H 2 div, = {v W ; div v W+1 }, v H 2 div, = v 2 W + div v 2 W +1 1/2. X 2 = H 2 curl, H 2 div,. v 2 W + div v 2 W +1 + curl v 2 W +1 1/2. These definitions will also be used with replaced by R 3. The argument used by Hanouzet [11] to prove the denseness of D in W m,2 can be easily adapted to establish that D is dense in the space H 2 div, and in the space H 2 curl, and so in X2. Therefore, denoting by n the exterior unit normal to the boundary, the normal trace v n and the tangential trace v n can be defined respectively in H 1/2 for the functions of H 2 div, and in H 1/2 for functions of H 2 curl,, where H 1/2 denotes the dual space of H 1/2. They satisfy the trace theorems; i.e, there exists a constant C such that v H 2 div,, v n H 1/2 C v H 2 div,, 1.6 v H 2 curl,, v n H 1/2 C v H 2 curl, 1.7 and the following Green s formulas holds: For any v H 2 div, and ϕ W v n, ϕ = v ϕ dx + ϕ div v dx, 1.8 where, denotes the duality pairing between H 1/2 and H 1/2. For any v H 2 curl, and ϕ W v n, ϕ = v curl ϕ dx curl v ϕdx, 1.9 where, denotes the duality pairing between H 1/2 and H 1/2. Remar 1.2. If v belongs to H 2 div, for some integer 1, then div v is in L 1 and Green s formula 1.8 yields v n, 1 = div v dx 1.10 But when 0, then div v is not necessarily in L 1 and 1.10 is generally not valid. Note also that when 0, W 1 does not contain the constants.

4 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 The closures of D in H 2 div, and in H2 curl, are denoted respectively by H 2 curl, and H 2 div, and can be characterized respectively by H 2 curl, = {v H 2 curl, : v n = 0 on }, H 2 div, = {v H 2 div, : v n = 0 on }. Their dual spaces are characterized by the following propositions: Proposition 1.3. A distribution f belongs to [ H 2 div, ] if and only if there exist ψ W and χ W 1, such that f = ψ + grad χ. Moreover f [ H2 div,] = max{ ψ W, χ W 1 }. 1.11 Proof. Let ψ W and χ W 1, we have v D, ψ + grad χ, v D D = ψ v χ div v dx. Therefore, the linear mapping l : v ψ v χ div vdx defined on D is continuous for the norm of H 2 div,. Since D is dense in H 2 div,, l can be extended by continuity to a mapping still called l [ H 2 div, ]. Thus ψ + grad χ is an element of [ H 2 div, ]. Conversely, Let E = W W+1 equipped by the following norm v E = v 2 W + div v 2 W +1 1/2. The mapping T : v H 2 div, v, div v E is an isometry from H 2 div, in E. Suppose G = T H 2 div, with the E-topology. Let S = T 1 : G H 2 div,. Thus, we can define the following mapping: v G f, Sv [ H2 div,] H 2 div, for f [ H 2 div, ] which is a linear continuous form on G. Thans to Hahn-Banach s Theorem, such form can be extended to a linear continuous form on E, denoted by Υ such that Υ E = f [ H2 div,]. 1.12 From the Riesz s Representation Lemma, there exist functions ψ W and χ W 1, such that for any v = v 1, v 2 E, Υ, v E E = v 1 ψ dx + v 2 χ dx, with Υ E = max{ ψ W, χ W 1 }. In particular, if v = T ϕ G, where ϕ D, we have f, ϕ [ H2 div, ] H 2 div, = ψ χ, ϕ [ H 2 div, ] H 2 div,, and 1.11 follows imeddiatly from 1.12. We sip the proof of the following result as it is similar to that of Proposition 1.3.

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 5 Proposition 1.4. A distribution f belongs to [ H 2 curl, ] if and only if there exist functions ψ W and ξ W 1, such that f = ψ + curl ξ. Moreover f [ H2 curl,] = max{ ψ W, ξ W 1 }. Definition 1.5. Let X 2,N, X2,T and X 2 be the following subspaces of X 2 : X 2,N = {v X 2 ; v n = 0 on }, X 2,T = {v X 2 ; v n = 0 on }, X 2 = X 2,N X 2,T. 2. Preliminary results Now, we give some results related to the Dirichlet problem and Neumann problem which are essential to ensure the existence and the uniqueness of some vectors potentials and one usually forces either the normal component to vanish or the tangential components to vanish. We start by giving the definition of the ernel of the Laplace operator for any integer Z: A 1 = {χ W : χ = 0 in and χ = 0 on }. In contrast to a bounded domain, the Dirichlet problem for the Laplace operator with zero data can have nontrivial solutions in an exterior domain; it depends upon the exponent of the weight. The result that we state below is established by Giroire in [10]. Proposition 2.1. For any integer 1, the space A 1 is a subspace of all functions in W of the form vp p, where p runs over all polynomials of and vp is the unique solution in W0 of the Dirichlet problem P 1 vp = 0 in and vp = p on. 2.1 The space A 1 is a finite-dimentional space of the same dimension as P 1 and A 1 = {0} when 0. Our second proposition is established also by Giroire in [10], it characterizes the ernel of the Laplace operator with Neumann boundary condition. For any integer Z, N 1 = {χ W χ : χ = 0 in and = 0 on }. n Proposition 2.2. For any integer 1, N 1 the subspace of all functions in W of the form wp p, where p runs over all polynomials of P 1 and wp is the unique solution in W 0 of the Neumann problem wp = 0 in and wp n = p n on. 2.2 Here also, we set N 1 = {0} when 0; N 1 is a finite-dimentional space of the same dimension as P 1 and in particular, N 0 = R.

6 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 Next, the uniqueness of the solutions of Problem 1.1 and Problem 1.2 will follow from the characterization of the ernel. For all integers in Z, we define Y 2,N = {w X 2,N : div w = 0 and curl w = 0 in } Y 2,T = {w X 2,T : div w = 0 and curl w = 0 in }. The proof of the following propositions can be easily deduced from [7]. Proposition 2.3. Let Z and suppose that is of class C 1,1, simply-connected and with a Lipshitz-continuous and connected boundary. If < 1, then Y,N 2 = {0}. If 1, then Y,N 2 = { vp p, p P 1 }, where vp is the unique solution in W 0 of the Dirichlet problem 2.1. Proof. Let Z and let w X 2,N such that div w = 0 and curl w = 0 in. Then since is simply-connected, there exists χ W, unique up to an additive constant, such that w = χ. But w n = 0, hence, χ is constant on is a connected boundary and we choose the additive constant in χ so that χ = 0 on. Thus χ belongs to A 1. Due to Proposition 2.1, we deduce that if < 1, χ is equal to zero and if 1, χ = vp p, where p runs over all polynomials of P 1 and vp is the unique solution in W0 of problem 2.1 and thus w = vp p. Now, to finish the proof we shall prove that vp p belongs to Y,N 2 and this is a simple consequence of the definition of p and vp. We sip the proof of the following result as it is entirely similar to that of Proposition 2.3. Proposition 2.4. Let the assumptions of Proposition 2.4 hold. If < 1, then Y,T 2 = {0}. If 1, then Y,T 2 = { wp p, p P 1 }, where wp is the unique solution in W 0 of the Neumann problem 2.2 The imbedding results that we state below are established by Girault in [7]. The first imbedding result is given by the following theorem. Theorem 2.5. Let 2 and assume that is of class C 1,1. Then the space X 2 1,T is continuously imbedded in W. In addition there exists a constant C such that for any ϕ X 2 1,T, ϕ W C ϕ W 1 + div ϕ W + curl ϕ W. 2.3 If in addition, is simply-connected, there exists a constant C such that for all ϕ X 2 1,T we have ϕ W C div ϕ W + curl ϕ W N + j=2 ϕ wq j dσ, 2.4 where {q j } N j=2 denotes a basis of {q P : q0 = 0}, N denotes the dimension of P and wq j is the corresponding function of N. Thus, the

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 7 seminorm in the right-hand side of 2.4 is a norm on X 2 1,T equivalent to the norm ϕ W. The second imbedding result is given by the following theorem. Theorem 2.6. Let 2 and assume that is of class C 1,1. Then the space X 2 1,N is continuously imbedded in W. In addition there exists a constant C such that for any ϕ X 2 1,N, ϕ W C ϕ W 1 + div ϕ W + curl ϕ W. 2.5 If in addition, is simply-connected and its boundary is connected, there exists a constant C such that for all ϕ X 2 1,N we have ϕ W C div ϕ W + curl ϕ W N + ϕ ndσ + ϕ nq j dσ, j=1 2.6 where the term ϕ ndσ can be dropped if 1 and where {q j} N j=1 denotes a basis of P. In other words, the seminorm in the right-hand side of 2.6 is a norm on X 2 1,N equivalent to the norm ϕ W. Finally, let us recall the abstract setting of Babuša-Brezzi s Theorem see Babuša [5], Brezzi [6] and Amrouche-Selloula [4]. Theorem 2.7. Let X and M be two reflexive Banach spaces and X and M their dual spaces. Let a be the continuous bilinear form defined on X M, let A LX; M and A LM; X be the operators defined by v X, w M, av, w = Av, w = v, A w and V = er A. The following statements are equivalent: i There exist β > 0 such that inf w M, w 0 sup v X, v 0 av, w v X w M β. 2.7 ii The operator A : X/V M is an isomophism and 1/β is the continuity constant of A 1. iii The operator A : M X V is an isomophism and 1/β is the continuity constant of A 1. Remar 2.8. As consequence, if the inf-sup condition 2.7 is satisfied, then we have the following properties: i If V = {0}, then for any f X, there exists a unique w M such that v X, av, w = f, v and w M 1 β f X. 2.8 ii If V {0}, then for any f X, satisfying the compatibility condition: v V, f, v = 0, there exists a unique w M such that 2.8. iii For any g M, there exists v X, unique up an additive element of V, such that: w M, av, w = g, w and v X/V 1 β g M.

8 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 3. Inequalities and inf-sup conditions In this sequel, we prove some imbedding results. More precisely, we show that the results of Theorem 2.5 and the result of Theorem 2.6 can be extended to the case where the boundary conditions v n = 0 or v n = 0 on are replaced by inhomogeneous one. Next, we study some problems posed in an exterior domain which are essentials to prove the regularity of solutions for Problem 1.1 and Problem 1.2. For any integer in Z, we introduce the following spaces: and Z 2,T = {v X 2 and v n H 1/2 }, Z 2,N = {v X 2 and v n H 1/2 } M 2,T = {v W +1, div v W+2, curl v W +2 and v n H 3/2 }. Proposition 3.1. Let = 1 or = 0, then the space Z 2,T is continuously imbedded in W +1 and we have the following estimate for any v in Z2,T : v W +1 C v W + curl v W +1 + div v W +1 + v n H 1/2. 3.1 Proof. Let = 1 or = 0 and let v any function of Z 2,T. Let us study the Neumann problem χ = div v in and n χ = v n on. 3.2 It is shown in [7, Theorems 3.7 and 3.9], that Problem 3.2 has a unique solution χ in W +1 /R if = 1 and χ is unique in W+1 if = 0. With the estimate χ W +1 C div v W +1 + v n H 1/2. 3.3 Let w = v grad χ, then w is a divergence-free function. Since W +1 W, then w X2,T. Applying Theorem 2.5, we have w belongs to W +1 and then v is in W +1. According to Inequality 2.3, we obtain w W +1 C w W + curl w W +1. Then, inequality 3.1 follows directly from 3.3. Similarly, we can prove the following imbedding result. Proposition 3.2. Suppose that is of class C 2,1. Then the space M 2 1,T is continuously imbedded in W 1 and we have the following estimate for any v in M 2 1,T : v W 1 C v W 0 + curl v W 1 + div v W 1 + v n H 3/2 3.4 Proof. Proceeding as in Proposition 3.1. Let v in M 2 1,T. Since is of class C 2,1, then according to [7, Theorem 3.9], there exists a unique solution χ.

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 9 in W 3,2 1 /R of Problem 3.2. Setting w = v grad χ. Since W 1 is imbedded in W 0, it follows from [7, Corollary 3.16], that w belongs to W 1 and moreover we have the estimate w W 1 C w W 0 + curl w W 1. Then v = w + grad χ belongs to W 1 and we have the estimate 3.4. Although we are under the Hilbertian case but the Lax-Milgram lemma is not always valid to ensure the existence of solutions. Thus, we shall establish two infsup conditions in order to apply Theorem 2.7. First recall the following spaces for all integers Z: { V,T 2 = z X 2,T : div z = 0 in and } z wq q dσ = 0, wq q N 1 and V 2,N = {z X 2,N : div z = 0 in and The first inf-sup condition is given by the following lemma. z nq dσ = 0, q P 1}. Lemma 3.3. The following inf-sup Condition holds: there exists a constant β > 0, such that curl ψ curl ϕ dx inf sup β. 3.5, ϕ 0 ψ X 2 2,T ϕ X 2 0,T ϕ V 2 0,T ψ V 2,T 2, ψ 0 Proof. Let g W 1 and let us introduce the Dirichlet problem χ = div g in, χ = 0 on. It is shown in [7, Theorem 3.5], that this problem has a solution χ unique up to an element of A 0 and we can choose χ such that W 1 χ W 1 C g W 1. Set z = g χ. Then we have z W 1, div z = 0 and we have z W 1 C g W 1. 3.6 Let ϕ any function of V0,T 2, by Theorem 2.5 we have ϕ X2 0,T Then due to 2.4 we can write ϕ X 2 0,T C curl ϕ W 1 = C sup g W 1, g 0 W 1. curl ϕ g dx. 3.7 g W 1 Using the fact that curl ϕ H 2 1div, and applying 1.8, we obtain curl ϕ χ dx = 0. 3.8 Now, let λ W 0 the unique solution of the problem λ = 0 in and λ = 1 on.

10 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 It follows from [7, Lemma 3.11] that λ n dσ = C 1 > 0. Now, setting z = z 1 C 1 z n, 1 λ. It is clear that z W 1, div z = 0 in and that z n, 1 = 0. Due to [7, Theorem 3.15], there exists a potential vector ψ W 1 such that and we have z = curl ψ, div ψ = 0 in and ψ n = 0 on. 3.9 vq N 1, In addition, we have the estimate ψ vq dσ = 0. 3.10 ψ W 1 C z W 1 C z W 1. 3.11 Using 3.10, we obtain that ψ belongs to V 2,T 2. Since ϕ is H1 in a neighborhood of, then ϕ has an H 1 extension in denoted by ϕ. Applying Green s formula in, we obtain 0 = divcurl ϕ dx = curl ϕ n, 1 = curl ϕ n, 1. Using the fact that curl ϕ in H 2 1div, and λ in W 1 and applying 1.8, we obtain 0 = curl ϕ n, 1 = curl ϕ n, λ = curl ϕ λ dx. 3.12 Using 3.8 and 3.12, we deduce that curl ϕ g dx = curl ϕ z dx = curl ϕ z dx. 3.13 From 3.11, 3.6 and 3.13, we deduce that curl ϕ g dx C curl ϕ z dx = C curl ϕ curl ψ dx. g W 1 z W 1 curl ψ W 1 Applying again 2.4 and using 3.10, we obtain curl ϕ g dx C curl ϕ curl ψ dx, g W 1 ψ X 2 2,T and the inf-sup Condition 3.5 follows immediately from 3.7. The second inf sup condition is given by the following lemma: Lemma 3.4. The following inf-sup Condition holds: there exists a constant β > 0, such that curl ψ curl ϕ dx inf sup β. 3.14, ϕ 0 ψ X 2 0,N ϕ X 2 2,N ϕ V 2 2,N ψ V0,N 2, ψ 0

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 11 Proof. The proof is similar to that of Lemma 3.3. Let g W 1 and let us introduce the generalized Neumann problem div χ g = 0 in and χ g n = 0 on. 3.15 It follows from [10] that Problem 3.15 has a solution χ W 1 and we have χ W 1 C g W 1. Setting z = g χ, then we have z H 2 1div, and div z = 0 with the estimate z W 1 C g W 1. 3.16 Let ϕ be any function of V 2,N 2. Due to Theorem 2.6, we have X2 2,N W 1 and by 2.6 we can write ϕ X 2 2,N C curl ϕ W 1 = C g W 1 sup, g 0 curl ϕ g dx. 3.17 g W 1 Observe that curl ϕ belongs to H 2 1div, with ϕ n = 0 on and χ W 1. Then using 1.8, we obtain curl ϕ χ dx = curl ϕ n, χ = 0. 3.18 Due to [7, Proposition 3.12], there exists a potential vector ψ W 1 such that In addition, we have z = curl ψ, div ψ = 0 in and ψ n = 0 on, 3.19 ψ n dσ = 0. 3.20 ψ W 1 C z W 1. 3.21 Then, we deduce that ψ belongs to V0,N 2. Using 3.16, 3.18 and 3.19, we deduce that curl ϕ g dx C g W 1 curl ϕ z dx = C z W 1 curl ϕ curl ψ dx Applying again 2.6 and using 3.20, we obtain curl ϕ g dx C curl ϕ curl ψ dx, g W 1 ψ X 2 0,N and the inf-sup condition 3.14 follows immediately from 3.17. curl ψ W 1. 4. Elliptic problems with different boundary conditions Next, we study the problem ξ = f and div ξ = 0 in, ξ n = g n on and ξ nq dσ = 0, q P. 4.1

12 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 Proposition 4.1. Let = 1 or = 0 and suppose that g n = 0 and let f [ H 2 1 curl, ] with div f = 0 in and satisfying the compatibility condition v Y1,N 2, f, v [ H2 1 curl ] H 2 1 curl = 0. 4.2 Then, Problem 4.1 has a unique solution in W and we have Moreover, if f in W +1 and is and satisfies the estimate W +1 ξ W C f [ H 2 1 curl,]. 4.3 of class C 2,1, then the solution ξ is in ξ W +1 C f W +1. 4.4 Proof. i On the one hand, observe that Problem 4.1 is reduced to the variational problem: Find ξ V 1,N 2 such that ϕ X 2 1,N, curl ξ curl ϕ dx = f, ϕ, 4.5 where the duality on is, =, [ H2 1 curl, ] H 2 1 curl,. On the other hand, 4.5 is equivalent to the problem: Find ξ V 1,N 2 such that ϕ V 1,N 2, curl ξ curl ϕ dx = f, ϕ. 4.6 Indeed, every solution of 4.5 also solves 4.6. Conversely, assume that 4.6 holds, and let ϕ X 2 1,N. Let us solve the exterior Dirichlet problem χ = div ϕ in and χ = 0 on. 4.7 It is shown in [7, Theorem 3.5] that problem 4.7 has a unique solution χ W /A. First case. if = 0, we set ϕ = ϕ χ 1 C 1 ϕ χ, 1 v1 1, where v1 is the unique solution in W 0 of the Dirichlet problem 2.1 and v1 C 1 = n dσ. It follows from [7, Lemma 3.11] that C 1 > 0 and since v1 1 belongs to Y1,N 2, we deduce that ϕ belongs to V2 1,N. Second case. if = 1, for each polynomial p in P 1, we tae ϕ of the form ϕ = ϕ χ vp p where vp is the unique solution in W 0 of the Dirichlet problem 2.1. The polynomial p is chosen to satisfy the condition ϕ n q dσ = 0 q P1. 4.8

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 13 To show that this is possible, let T be a linear form defined by T : P1 R 4, vp p vp p T p = dσ, x 1 dσ, n n vp p vp p x 2 dσ, x 3 dσ, n n where {1, x 1, x 2, x 3 } denotes a basis of P1. It is shown in the proof of [8, Theorem 7], that if vp p q dσ = 0 q P1, n then p = 0. This implies that T is injective and so bijective. And so, there exists a unique p in P1 so that condition 4.8 is satisfied and since vp p belongs to Y 2 2,N, we prove that ϕ V2 2,N. Finally, using 4.2, we obtain for = 0 and = 1 that f, vp p = 0 and f, v1 1 = 0 and as D is dense in H 2 1 curl,, we obtain that Then we have curl ξ curl ϕ dx = f, χ = 0. curl ξ curl ϕ dx = f, ϕ. Then Problem 4.5 and Problem 4.6 are equivalent. Now, to solve Problem 4.6, we use Lax-Milgram lemma for = 0 and the inf-sup condition 3.14 for = 1. Let us start by = 0. We consider the bilinear form a : V 1,N 2 V2 1,N R such that aξ, ϕ = curl ξ curl ϕ dx. According to Theorem 2.6, a is continuous and coercive on V 1,N 2. Due to Lax- Milgram lemma, there exists a unique solution ξ V 1,N 2 of Problem 4.6. Using again Theorem 2.6, we prove that this solution ξ belongs to W 0 and the following estimate follows immediately ξ W 0 C f [ H 2 1 curl,]. 4.9 When = 1, we have that Problem 4.6 satisfies the inf-sup condition 3.14. Let us consider the mapping l : V 2,N 2 R such that lϕ = f, ϕ. It is clear that l belongs to V 2,N 2 and according to Remar 2.8, there exists a unique solution ξ V0,N 2 of Problem 4.6. Due to Theorem 2.6, we prove that this solution ξ belongs to W 1. It follows from Remar 2.8 i that ξ W 1 C f [ H 2 2 curl,]. 4.10 ii We suppose in addition that f is in W +1 for = 1 or = 0 and is of class C 2,1 and we set z = curl ξ, where ξ W is the unique solution of Problem 4.1. Then we have z W, curl z = f W +1, div z = 0 and z n = 0 on

14 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 and thus z belongs to X 2,T. By Theorem 2.5, we prove that z belongs to W +1 and using 2.4, we prove that z satisfies z W +1 C f W +1. 4.11 As a consequence ξ satisfies ξ W, curl ξ W +1, div ξ = 0 and ξ n = 0 on. Applying [7, Corollary 3.14], we deduce that ξ belongs to W +1 and using in addition the boundary condition of 4.1 we prove that ξ W +1 C curl ξ W +1. 4.12 Finally, estimate 4.4 follows from 4.11 and 4.12. Corollary 4.2. Let = 1 or = 0 and let f [ H 2 1 curl, ] with div f = 0 in and g H 1/2 and satisfying the compatibility condition 4.2. Then, Problem 4.1 has a unique solution ξ in W and we have: ξ W f C [ H2 1 curl,] + g n H 1/2. 4.13 Moreover, if f in W +1, g in H 3/2 and is of class C 2,1, then the solution ξ is in W +1 and satisfies ξ W +1 C f W +1 + g n H 3/2. 4.14 Proof. Let = 0 or = 1 and let g H 1/2. We now that there exists ξ 0 in H 1 with compact support satisfying ξ 0 = g τ on and div ξ 0 = 0 in, where g τ is the tangential component of g on. Since support of ξ 0 is compact, we deduce that ξ 0 belongs to W for = 1 or = 0 and satisfies ξ 0 W C g τ H 1/2. 4.15 Setting z = ξ ξ 0, then Problem 4.1 is equivalent to: find z W such that z = f + ξ 0 and div z = 0 in, z n = 0 on and z n q dσ = 0, q P 4.16. Observe that F = f curl curl ξ 0 belongs to [ H 2 1 curl, ]. Since D is dense in H 2 1 curl,, we have for any v Y2 1,N : curl curl ξ 0, v = curl ξ 0 curl v dx = 0. Thus F satisfies the compatibility condition 4.2. Due to Proposition 4.1, there exists a unique z W solution of problem 4.16 such that z W C F [ H C f 2 1 curl,] [ H2 1 curl,] + curl ξ 0 W. 4.17 Then ξ = z + ξ 0 belongs to W is the unique solution of 5.8 and estimate 4.13 follows immediately from 4.15 and 4.17.

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 15 Regularity of the solution: Suppose in addition that is of class C 2,1, f in W +1 and g in H 3/2. Then the function ξ 0 defined above belongs to H 2 with compact support and thus ξ 0 belongs to W +1 and we have ξ 0 W +1 C g τ H 3/2. 4.18 Using again Proposition 4.1, we prove that z belongs to W +1 and satisfies z W +1 C F W +1. Then ξ is in W +1 and estimate 4.14 follows from 4.18. The next theorem solves an other type of exterior problem. Theorem 4.3. Let = 1 or = 0 and let v belongs to W. Then, the following problem ξ = curl v and div ξ = 0 in, ξ n = 0 and curl ξ v n = 0 on, ξ wq q dσ = 0, wq q N has a unique solution ξ in W and we have 4.19 ξ W C v W. 4.20 Moreover, if v W +1 and is of class C 2,1, then the solution ξ is in W and satisfies the estimate +1 ξ W +1 C v W +1. 4.21 Proof. At first observe that if ξ W is a solution of Problem 4.19 for = 1 or = 0, then curl ξ v belongs to H 2 curl, and thus curl ξ v n is well defined in and belongs to H 1/2. On the other hand, note that 4.19 can be reduced to the variational problem: Find ξ V 1,T 2 such that ϕ X 2 1,T curl ξ curl ϕ dx = v curl ϕ dx. 4.22 Indeed, every solution of 4.19 also solves 4.22. Conversely, let ξ V 2 1,T a solution of the problem 4.22. Then, Then ϕ D, curl curl ξ curl v, ϕ D D = 0. ξ = curl v in. 4.23 Moreover, by the fact that ξ belongs to the space V 1,T 2 we have div ξ = 0 in and ξ n = 0 on. Then, it remains to verify the boundary condition curl ξ v n = 0 on. Now setting z = curl ξ v, then z belongs to H 2 curl,. Therefore, 4.23 becomes curl z = 0 in. 4.24

16 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 Let ϕ X 2 1,T, by Theorem 2.5 we have X2 1,T W. Than s to 1.9 we obtain z curl ϕ dx = z n, ϕ H 1/2 H 1/2 + curl z ϕ dx. 4.25 Compare 4.25 with 4.22 and using 4.24, we deduce that ϕ X 2 1,T, z n, ϕ = 0. Let now µ any element of the space H 1/2. As is bounded, we can fix once for all a ball B R, centered at the origin and with radius R, such that B R. Setting R = B R, then we have the existence of ϕ in H 1 R such that ϕ = 0 on B R and ϕ = µ t on, where µ t is the tangential component of µ on. The function ϕ can be extended by zero outside B R and the extended function, still denoted by ϕ, belongs to W 1,p α, for any α since its support is bounded. Thus ϕ, belongs to and W. It is clear that ϕ belongs to X2 1,T z n, µ = z n, µ t = z n, ϕ = 0. 4.26 This implies that z n = 0 on which is the last boundary condition in 4.19. On the other hand, let us introduce the problem: Find ξ V 1,T 2 such that ϕ V 1,T 2 curl ξ curl ϕ dx = v curl ϕ dx. 4.27 Problem 4.27 can be solved by Lax-Milgram lemma if = 0 and by Lemma 3.3 if = 1. We start by the case = 1. Observe that Problem 4.27 satisfies the inf-sup condition 3.5. Let consider the mapping l : V0,T 2 R such that lϕ = v curl ϕ dx. It is clear that l belongs to V2 0,T and according to Remar 2.8, there exists a unique solution ξ V 2,T 2. Applying Theorem 2.5, we deduce that this solution ξ belongs to W 1. It follows from Remar 2.8 i and Theorem 2.6 that ξ W 1 C l V0,T 2 C v W 1. 4.28 For = 0, let us consider the bilinear form b : V 1,T 2 V2 1,T R such that bξ, ϕ = curl ξ curl ϕ dx. According to Theorem 2.5, b is continuous and coercive on V 1,T 2. Due to Lax- Milgram lemma, there exists a unique solution ξ V 1,T 2 of Problem 4.27. Using again Theorem 2.5, we prove that this solution ξ belongs to W 0 and estimate 4.20 follows immediately. Next, we extend 4.27 to any test function in X 2 1,T. Let ϕ X2 1,T and let us solve the exterior Neumann problem χ χ = div ϕ in and = 0 on. 4.29 n It is shown in [7, Lemma 3.7 and Theorem 3.9] that this problem has a unique solution χ in W 1 if = 1 and unique up to a constant if = 0. Set ϕ = ϕ χ. 4.30

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 17 It is clear that for = 0 and = 1, ϕ wq q dσ = 0 for any wq q N. Then ϕ belongs to V2 1,T. Now, if 4.27 holds, we have curl ξ curl ϕ dx = curl ξ curl ϕ dx = v curl ϕ dx. Hence, problem 4.22 and problem 4.27 are equivalent. This implies that problem 4.19 has a unique solution ξ in W for = 0 or = 1. Regularity. Now, we suppose that v W +1 W and is of class C 2,1. Let ξ W the wea solution of 4.19 and we set z = curl ξ v. It is clear that z belongs to X 2,N. Applying Theorem 2.6, we obtain that z W +1 and using 2.5 and 4.20 we obtain that z W +1 C z W + div z W C C v W. +1 This implies that ξ satisfies curl ξ W + v W + div v W +1 +1 4.31 ξ W, div ξ = 0 W +1, curl ξ W +1, ξ n = 0 on. Applying [7, Corollary 3.16], we deduce that ξ belongs to W +1 and using 4.31, we obtain ξ W +1 C ξ W + curl ξ W +1 C v W + z W +1 + v W +1 v W. +1 This completes the proof of the theorem. As consequence, we can prove other imbedding results. We start by the following theorem. Theorem 4.4. Let = 1 or = 0. Then the space Z 2,N is continuously imbedded in W +1 and we have the following estimate for any v in Z2,N : v W +1 C v W + curl v W +1 + div v W +1 + v n H 1/2. 4.32 Proof. Let = 1 or = 0 and let v be any function of Z 2,N. We set z = curl ξ v where ξ W is the solution of the problem 4.19. Hence, z belongs to the space X 2,N. By Theorem 2.6 and 2.5, z even belongs to W +1 with the estimate z W +1 C z W + div z W +1 + curl z W +1. 4.33 Then, it suffices to prove that curl ξ W +1 to obtain v W +1. Setting ω = curl ξ. It is clear that ω n dσ = 0 4.34

18 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 and then ω satisfies ω = curl curl v and div ω = 0 in ω n = v n on and ω nq dσ = 0, q P 1. 4.35 Note that curl v W +1 then curl curl v is in [ H 2 2 curl, ] and we have v n H 1/2. Since D is dense in H 2 2 curl,, we prove that φ Y+2,N 2, curl curl v, φ [ H2 2 curl,] H 2 2 curl, = 0. Due to Corollary 4.2, the function ω belongs to W +1 and satisfies the estimate ω W +1 C curl curl v [ H2 2 curl,] + v n H 1/2 C curl v W +1 + v n H 1/2. 4.36 Finally, estimate 4.32 can be deduced by using inequalities 4.33 and 4.36. Before giving the second imbedding result, we need to introduce the following space for any integer in Z, { M 2,N = v W +1, div v W+2, curl v W +2, } v n H 3/2. Proposition 4.5. Suppose that is of class C 2,1. Then the space M 2 1,N is continuously imbedded in W 1 and we have the following estimate for any v in M 2 1,N, v W 1 C v W 0 + curl v W 1 + div v W 1 + v n H 3/2 4.37 Proof. The proof is very similar to that of Theorem 4.4. Let v be any function of M 2 1,N and set z = curl ξ v where ξ W 0 is the solution of the problem 4.19. According to [7, Corollary 3.14], we prove that z belongs to W 1 with the estimate z W 1 C z W 0 + div z W 1 + curl z W 1. 4.38 Then, it suffices to prove that curl ξ W 1 to obtain v W 1. We set ω = curl ξ. Using [7, Theorem 3.1], we prove that ω satisfies Problem 4.35. Using the regularity of Corollary 4.2, we prove that ω belongs to W 1 and satisfies ω W C curl curl v W 1 + v n H 3/2 and then estimate 4.37 follows from 4.38. 5. Generalized solutions for 1.1 and 1.2 We start this sequel by introducing the space E 2 = {v W 0 : v [ H 2 1div, ] }. This is a Banach space with the norm v E2 = v W 0 + v [ H 2 1 div, ]..

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 19 We have the following preliminary result. Lemma 5.1. The space D is dense in E 2. Proof. Let P be a continuous linear mapping from W 0 to W 0 R3, such that P v = v and let l E 2, such that for any v D, we have l, v = 0. We want to prove that l = 0 on E 2. Then there exists f, g W 0 R 3 H 2 1div, such that: for any v E 2, l, v = f, P v W 0 R 3 W 0 R3 + v, g [ H 2 1 div, ] H 2 1 div,. Observe that we can easily extend by zero the function g in such a way that g H 2 1div, R 3. Now we tae ϕ DR 3. Then we have by assumption that f, ϕ W 0 R 3 W 0 R3 + g ϕdx = 0, R 3 because f, ϕ = f, P v where v = ϕ. Thus we have f + g = 0 in D R 3. Then we can deduce that g = f W 0 R 3 and due to [2, Theorem 1.3], there exists a unique λ W 0 R3 such that λ = g. Thus the harmonic function λ g belonging to W 1 R3 is necessarily equal to zero. Since g W 0 and g W 0 R3, we deduce that g W 0. As D is dense in W 0, there exists a sequence g D such that g g in W 0, when. Then g g in L 2. Since W 0 is imbedded in W 1, we deduce that g g in H 2 1div,. Now, we consider v E 2 and we want to prove that l, v = 0. Observe that: l, v = g, P v W 0 R 3 W 0 R3 + v, g [ H 2 1 div,] H 2 1 div, = lim g vdx + v, g [ H2 1 div, ] H 2 1 div, = lim g vdx + v g dx = 0. As a consequence, we have the following result. Corollary 5.2. The linear mapping γ : v curl v n defined on D can be extended to a linear continuous mapping γ : E 2 H 1/2. Moreover, we have the Green formula:for any v E 2 and any ϕ W 0 such that div ϕ = 0 in and ϕ n = 0 on, v, ϕ [ H2 1 div,] H 2 1 div, = curl v curl ϕ dx curl v n, ϕ, 5.1 where the duality on is defined by, =, H 1/2 H 1/2. Proof. Let v D. Observe that if ϕ W 0 such that ϕ n = 0 on we deduce that ϕ X 2 1,T, then 5.1 holds for such ϕ. Now, let µ H1/2, then there exists ϕ W 0 such that ϕ = µ t on and that div ϕ = 0 with ϕ W 0 C µ t H 1/2 C µ H 1/2. 5.2

20 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 As a consequence, using 5.1, we have Thus, curl v n, µ C v E 2 µ H 1/2. curl v n H 1/2 C v E2. We deduce that the linear mapping γ is continuous for the norm E 2. Since D is dense in E 2, γ can be extended to by continuity to γ LE 2, H 1/2 and formula 5.1 holds for all v E 2 and ϕ W 0 such that div ϕ = 0 in and ϕ n = 0 on. Proposition 5.3. Let f belongs to W 1 with div f = 0 in, g H1/2 and h H 1/2 verify the compatibility conditions: for any v Y1,T 2, f v dx + h n, v H 1/2 H 1/2 = 0, 5.3 f n + div h n = 0 on, 5.4 where div is the surface divergence on. Then, the problem z = f and div z = 0 in, z n = g and curl z n = h n on, has a unique solution z in W 0 satisfying the estimate z W 0 C f W 1 + g H 1/2 + h n H 1/2 5.5. 5.6 Moreover, if h in H 1/2, g in H 3/2 and is of class C 2,1, then the solution z is in W 1 and satisfies the estimate z W 1 C f W 1 + g H 3/2 + h n H 1/2. 5.7 Proof. First, note that if h H 1/2, then h n also belongs to H 1/2. On the other hand, let us consider the Neumann problem: θ θ = 0 in and = g on. 5.8 n It is shown in [7, Theorem 3.9], that this problem has a unique solution θ W 0 /R satisfying the estimate θ W 0 C g H 1/2. 5.9 Setting ξ = z θ, then problem 5.5 becomes: find ξ W 0 such that ξ = f and div ξ = 0 in, 5.10 ξ n = 0 and curl ξ n = h n on. Now, observe that problem 5.10 is reduced to the variational problem: Find ξ V 1,T 2 such that ϕ X 2 1,T curl ξ curl ϕ dx = f ϕ dx + h n, ϕ. 5.11 Indeed, every solution of 5.10 also solves 5.11. Conversely, let ξ a solution of the problem 5.11. Then, ϕ D, curl curl ξ f, ϕ D D = 0.

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 21 So ξ = f in. Moreover, by the fact that ξ belongs to the space V 1,T 2, we have div ξ = 0 in and ξ n = 0 on. Then, it remains to verify the boundary condition curl ξ n = h n on. Observe that ξ belongs to E 2 so by 5.1 and comparing with 5.11 we deduce that for any ϕ X 2 1,T, we have curl ξ n, ϕ = h n, ϕ. Proceeding as in the proof of Theorem 4.3, we prove that curl ξ n = h n on. On the other hand, let us introduce the following problem: Find ξ V 1,T 2 such that ϕ V 1,T 2 curl ξ curl ϕ dx = f ϕ dx + h n, ϕ. 5.12 As in the proof of Theorem 4.3, we use Lax-Milgram lemma to prove the existence of a unique solution ξ in V 1,T 2 of Problem 5.12. Using Theorem 2.5, we prove that this solution ξ belongs to W 0 and the following estimate follows immediately ξ W 0 C f W 1 + h n H 1/2. 5.13 Next, we extend 5.12 to any test function ϕ in X 2 1,T. Let ϕ X2 1,T and let us solve the exterior Neumann problem: χ χ = div ϕ in and = 0 on. 5.14 n It is shown in [10, Theorem 4.12] that this problem has a unique solution χ in W 0 up to an additive constant. Then, we set ϕ = ϕ χ. 5.15 Since W 0 is imbedded in W 1, then ϕ belongs to V2 1,T. Now, if 5.12 holds, we have curl ξ curl ϕ dx = f ϕ dx + h n, ϕ f χdx h n, χ. Using 1.8 and 5.4, we obtain curl ξ curl ϕ dx = f ϕ dx + h n, ϕ. This implies that problem 5.11 and problem 5.12 are equivalent and thus problem 5.10 has a unique solution ξ in W 0. Finally, we set z = ξ+ θ W 0 the unique solution of 5.5. Finally, 5.6 follows immediately from 5.13 and 5.9. Regularity of the solution. We suppose in addition that h is in H 1/2, g in H 3/2 and is of class C 2,1 and let z in W 0 be the wea solution of Problem 5.5. Setting ω = curl z, then ω satisfies ω L 2, div ω = 0 W 1, curl ω = f W 1, ω n = curl z n H1/2. Applying Theorem 4.4 with = 0, we prove that ω belongs to W 1. This implies that z satisfies z W 0, div z = 0 W1, curl z W 1 z n H3/2.

22 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 Applying Proposition 3.2, we prove that z belongs to W 1 and we have the estimate 5.7. Next, we solve the Stoes problem 1.1. Theorem 5.4 Wea solutions for 1.1. Suppose that g = 0 and χ = 0. For f given in [ H 2 1div, ] and h given in H 1/2 satisfying 5.3. The Stoes problem 1.1 has a unique solution u, π W 0 L2 and we have the estimate u W 0 + π L 2 C f [ H2 1 div,] + h n H 1/2. 5.16 Proof. At first, observe that problem 1.1 is reduced to the variational problem: Find u V 2 1,T such that ϕ V 2 1,T, curl u curl ϕ dx = f, ϕ [ H2 1 div, ] H 2 1 div, + h n, ϕ. 5.17 Indeed, every solution of 1.1 also solves 5.17. Conversely, let u a solution of problem 5.17. Then for all ϕ D such that div ϕ = 0, u f, ϕ D D = 0. By De Rham theorem, there exists q D such that u f = q in. Note that [ H 2 1div, ] is imbedded in W 0 and thus u f W 0. It follows from [7, Theorem 2.7], that there exists a unique real constant C and a unique π L 2 such that π has the decomposition q = π + C. Observe that since f and π are two elements of [ H 2 1div, ], it is the same for u. Since D is dense in H 2 1div,, we obtain for any ϕ H 2 1div, such that div ϕ = 0: π, ϕ [ H 1div, ] H 1div, = 0. Moreover, if ϕ V 1,T 2, using Corollary 5.2 we have u, ϕ [ H2 1 div, ] H 2 1 div, = curl u curl ϕ dx curl u n, ϕ H 1/2 H. 1/2 We deduce that for all ϕ V 2 1,T curl u n, ϕ H 1/2 H 1/2 = h n, ϕ H 1/2 H 1/2. Let now µ any element of the space H 1/2. So, there exists an element ϕ W 0 such that div ϕ = 0 in and ϕ = µ t on. It is clear that ϕ V 1,T 2 and curl u n, µ h n, µ = curl u n, µ t h n, µ t = curl u n, ϕ h n, ϕ = 0. This implies that cur u n = h n on. As a consequence, Problem 5.17 and 1.1 are equivalent. As in the proof of Theorem 4.3, we use Lax-Milgram lemma to prove the existence of a unique solution u in V 1,T 2 of Problem 5.17. Using

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 23 Theorem 2.5, we prove that this solution u belongs to W 0. Then the pair u, π W 0 L2 is the unique solution of the problem 1.1. The estimate 5.16 follows from 2.4. Corollary 5.5. Let f, χ, g, h such that f [ H 2 1div, ], χ L 2, g H 1/2, h H 1/2, and that 5.3 holds. Then, the Stoes problem 1.1 has a unique solution u, π W 0 L2 and we have: u W 0 + π L 2 C f [ H2 1 div, ] + χ L2 + g H 1/2 + h n H 1/2. 5.18 Proof. First case: We suppose that χ = 0. Let θ W 0 be a solution of the exterior Neumann problem 5.8. Setting z = u θ, then, problem 1.1 becomes: Find z, π W 0 L2 such that z + π = f and div z = 0 in, z n = 0 and curl z n = h n on. 5.19 Due to Theorem 5.4, this problem has a unique solution z, π W 0 L2. Thus u = z + θ belongs to W 0 and using 5.9 and 5.16, we deduce that u W 0 + π L 2 C f [ H2 1 div, ] + g H 1/2 + h n H 1/2. 5.20 Second case: We suppose that χ L 2. We solve the following Neumann problem in : θ θ = χ in, = g on. 5.21 n It follows from [7, Theorem 3.9] that Problem 5.21 has a unique solution θ in W 0 /R and we have θ W 0 /R C χ L 2 + g H 1/2. 5.22 Setting z = u θ, then Problem problem 1.1 becomes: Find z, π W 0 L 2 such that z + π = f + χ and div z = 0 in, z n = 0 and curl z n = h n on. 5.23 Observe that f+ χ belongs to [ H 2 1div, ] and χ, v [ H2 1 div, ] H 2 1 div, = 0 for all v in Y1,T 2. According to the first step, this problem has a unique solution z, π W 0 L2. Thus u = z + θ belongs to W 0 and estimate 5.18 follows from 5.20 and 5.22. Now, we study the problem 1.2. Theorem 5.6 Wea solutions for 1.2. Assume that χ = 0, f is in the space [ H 2 1curl, ], g is in H 1/2 and π 0 is in H 1/2, satisfying the compatibility condition λ Y 2 1,N, f, λ [ H2 1 curl,] H 2 1 curl, = λ n, π 0. 5.24

24 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 Then the Stoes problem 1.2 has a unique solution u, π W 0 W1 and we have u W 0 + π W 1 C f [ H 2 1 curl, ] + g n H 1/2 + π 0 H 1/2. Proof. First, we consider the problem 5.25 π = div f in, π = π 0 on. 5.26 Since f [ H 2 1curl, ], we deduce from Proposition 1.4 that div f belongs to W 1. Now, let v1 1 an element of A 0, it is clear that v1 1 belongs to Y1,N 2. Then using the density of D in W 1 and 5.24, we prove that div f, v1 1 W 1 W 1 = f, v1 1 [ H 2 1 curl,] H 2 1 curl, = v1 1 n, π 0. 5.27 Since 5.27 is satisfied, we apply [7, Theorem 3.6] to prove that Problem 5.26 has a unique solution π W 1 and we have the following estimate: π W 1 C div f W 1 + π 0 H 1/2. 5.28 Setting F = f π, then F belongs to [ H 2 1curl, ]. Thus Problem 1.2 becomes: Find u W 0 such that: u = F and div u = 0 in, 5.29 u n = g n on and u n dσ = 0. Using 5.24 and the fact that D is dense in H 2 1curl,, we prove that λ Y1,N 2, F, λ [ H2 1 curl,] H 2 1 curl, = 0. 5.30 Therefore, F satisfies the assumptions of Corollary 4.2 and thus Problem 5.29 has a unique solution u W 0 with u W F 0 C [ H2 1 curl,] + g n H 1/2. 5.31 Thus estimate 5.25 follows from 5.31 and from 5.28. Corollary 5.7. Let f, χ, g, π 0 such that f [ H 2 1curl, ], χ W 1, g H 1/2, π 0 H 1/2, and satisfying the compatibility condition: λ Y 2 1,N, f, λ [ H2 1 curl,] H 2 1 curl, = λ n, π 0 χ. 5.32 Then the Stoes problem 1.2 has a unique solution u, π W 0 W1. Moreover, we have the estimate u W 0 + π W 1 C f [ H2 1 curl, ] + g n H 1/2 + π 0 H 1/2 + χ W 1. 5.33

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 25 Proof. First, we consider the problem π = div f + χ in, π = π 0 on. 5.34 Since f [ H 2 1curl, ], we deduce from Proposition 1.4 that div f + χ belongs to W 1. Proceeding as in the proof of Theorem 5.6, we prove that div f + χ, v1 1 W 1 W 1 = v1 1 n, π 0 and then we apply [7, Theorem 3.6] to prove that Problem 5.34 has a unique solution π W 1 and we have the following estimate: π W 1 C div f + χ W 1 + π 0 H 1/2 Thus Problem 1.2 becomes: Find u W 0 such that u = f π and div u = χ in, u n = g n on and u n dσ = 0. On the other hand, let us solve the Dirichlet problem θ = χ in, θ = 0 on.. 5.35 5.36 Since W 1 is imbedded in L 2, it follows from [7, Theorem 3.5], that this problem has a unique solution θ W 0 there is no compatibility condition and we have the estimate Setting θ W 0 C χ L 2. 5.37 z = u θ 1 θ n, 1 v1 1, C 1 where v1 is the unique solution in W 0 of the Dirichlet problem 2.1 and C 1 = v1 n dσ. We now from [7, Lemma 3.11] that C 1 > 0 and that v1 1 belongs to Y1,N 2. Then Problem 5.36 becomes: Find z W 0 such that z = f π + χ and div z = 0 in, z n = g n on and z n dσ = 0. 5.38 Now, we will solve the Problem 5.38. Setting F = f π + χ, then F belongs to [ H 2 1curl, ]. Using 5.32 and the fact that D is dense in H 2 1curl,, we prove that λ Y1,N 2, F, λ [ H2 1 curl,] H 2 1 curl, = 0. 5.39 Therefore, F satisfies the assumptions of Corollary 4.2 and thus Problem 5.38 has a unique solution z W 0 with z W F 0 C [ H2 1 curl,] + g n H 1/2 5.40 and estimate 5.33 holds.

26 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 6. Strong solutions for 1.1 and 1.2 We prove in this sequel the existence and the uniqueness of strong solutions for Problem 1.1 and 1.2, we start with Problem 1.1. Theorem 6.1. Suppose that is of class C 2,1. Let f, χ, g, h be such that f W 1, χ W1, g H 3/2, h H 1/2, and that 5.3 holds. Then, the Stoes problem 1.1 has a unique solution u, π W1 and we have W 1 u W 1 + π W 1 f W 1 + χ W 1 + g H 3/2 + h n H 1/2. 6.1 Proof. First case: χ = 0. Since W 1 is included in [ H 2 1div, ], we deduce that we are under the hypothesis of Corollary 5.5 and so Problem 1.1 has a unique solution u, π W 0 L2. Setting z = curl u, then z satisfies z L 2, div z = 0 W 1, z = curl f, z n = h n H 1/2. Applying Corollary 4.2 with = 1 and the uniqueness argument, we prove that z belongs to W 1. This implies that u satisfies u W 0, div u = 0 W1, curl u W 1, u n = g H3/2. Applying Proposition 3.2, we prove that u belongs to W 1 and thus π = f + u W 1. Since π is in L2 then π is in W 1. Second case: χ is in W 1. Since is of class C 2,1, it follows from [7, Theorem 3.9] that there exists a unique solution θ in W 3,2 1 /R satisfies Problem 5.21 and θ W 3,2 1 /R C χ W 1 + g H 3/2 The rest of the proof is similar to that Corollary 5.5.. 6.2 Remar 6.2. Assume that the hypothesis of Theorem 6.1 hold and suppose in addition that χ = 0. Let u, π W 0 L2 the unique solution of Problem 1.1 then π satisfies the problem div π f = 0 in and π f n = div h n on. 6.3 It follows from [10] that Problem 6.3 has a solution π in W 1. Setting F = π f W 1. Then problem 1.1 becomes u = F and div u = 0 in, u n = g and curl u n = h n on. Therefore, F, g and h satisfy the assumptions of Proposition 5.3 and thus u belongs to W 1. Next, we study the regularity of the solution for Problem 1.2. Theorem 6.3. Suppose that is of class C 2,1. Let f, χ, g, π 0 be such that f W 1, χ W1, g H 3/2, π 0 H 1/2,

EJDE-2013/196 STOKES PROBLEM WITH SEVERAL BOUNDARY CONDITIONS 27 and satisfying the compatibility condition 5.32. Then the Stoes problem 1.2 has a unique solution u, π W 1 W1. Moreover, we have the estimate u W 1 + π W 1 C f W 1 + g n H 3/2 + π 0 H 1/2 + χ W 1. 6.4 Proof. First case: We suppose that χ = 0. Since W 1 is in [ H 2 1curl, ], we deduce that we are under the hypothesis of Corollary 5.6 and so Problem 1.2 has a unique solution u, π W 0 W1. Setting z = curl u. Observe that u n = g n belongs to H 3/2 and thus curl u n belongs to H 1/2 and so z satisfies z L 2, div z = 0 W 1, curl z = f π W 1, z n = curl u n H1/2. Applying Proposition 3.1 with = 0, we prove that z belongs to W 1. This implies that u satisfies u W 0, div u = 0 W1, curl u W 1, u n = g n H3/2. Applying Proposition 4.5, we prove that u belongs to W 1. Second case: χ is in W 1. The proof of this case is very similar to that Corollary 5.7. References [1] C. Amrouche, C. Bernardi, M. Dauge, V. Girault; Vector potentials in three-dimensional nonsmooth domains. Math. Meth. Appl. Sci 21, 823 864, 1998. [2] C. Amrouche, V. Girault and J. Giroire; Dirichlet and Neumann exterior problems for the n- dimensional laplace operator, an approach in weighted Sobolev spaces. J. Math. Pures Appl. 76 1, 55 81, 1997. [3] C. Amrouche, N. Seloula; Stoes equations and elliptic systems with nonstandard boundary conditions. C. R. Math. Acad. Sci. 349 11-12, 703 708, 2011. [4] C. Amrouche, N. Seloula; L p -Theory for vector potentials and Sobolev s inequalities for vector fields: Application to the Stoes equations with pressure boundary conditions. Mathematical Models and Methods in Applied Sciences. 23 1, 37 92, 2013. [5] I. Babuša; The finite element method with Lagrangian multipliers. Numer. Math. 20, 179 192, 1973. [6] F. Brezzi; On the existence, uniqueness and approximation of saddle-point problems arising from Lagrange multipliers. R. A. I. R. O. Anal. Numer. R2, 129 151, 1974. [7] V. Girault; The Stoes problem and vector potential operator in three-dimensional exterior domains: An approach in weighted Sobolev spaces. Differential and Integral Equations 7 2, 535 570, 1994. [8] V. Girault, J. Giroire, A. Sequeira; A stream-function-vorticity variational formulation for the exterior Stoes problem in weighted Sobolev spaces. Mathematical Methods in the Applied Sciences 15, 345 363, 1992. [9] V. Girault, A. Sequeira; A well posed problem for the exterior Stoes equations in two and three dimensions. Arch. Rat. Mech. Anal 114, 313 333, 1991. [10] J. Giroire; Etude de quelques problèmes aux limites extérieurs et résolution par équations intégrales. Thèse, Université Paris VI, 1987. [11] B. Hanouzet; Espaces de Sobolev avec poids, Application au problème de Dirichlet dans un demi-espace. Rend. Sem. Univ. Padova, 46, 227 272, 1971.

28 C. AMROUCHE, M. MESLAMENI EJDE-2013/196 Chérif Amrouche Laboratoire de Mathématiques et de leurs Applications, Pau - CNRS UMR 5142, Université de Pau et des Pays de l Adour, IPRA, Avenue de l Université - 64000 Pau - France E-mail address: cherif.amrouche@univ-pau.fr Mohamed Meslameni Laboratoire de Mathématiques et de leurs Applications, Pau - CNRS UMR 5142, Université de Pau et des Pays de l Adour, IPRA, Avenue de l Université - 64000 Pau - France E-mail address: mohamed.meslameni@univ-pau.fr